NO - Summary of Chapter 3 (Part I)

Given the constraints of a LP:

$$\begin{array}{rl}& \overline{\overline{A}}\cdot \overline{x}=\overline{b}\\ & \overline{x}\ge 0\end{array}$$

we can divide the matrix $\overline{\overline{A}}$ and the vector $\overline{b}$ in bases and non-bases, where:

A base $B$ is a set of linearly independent $m$ columns of $\overline{\overline{A}}$

So we can always write

$$\overline{\overline{A}}=\left[\overline{\overline{B}},\overline{\overline{N}}\right]$$

where: $\overline{\overline{B}}$ is the base of $\overline{\overline{A}}$ and $\overline{\overline{N}}$ is the non-base

So we have separate $x$ in the same way:

$$\overline{x}=\left[{\overline{x}}_B,{\overline{x}}_N\right]$$

where: ${\overline{x}}_B$ (basic variables) is the part of $\overline{x}$ that multiplies $\overline{\overline{B}}$, and ${\overline{x}}_N$ (non-basic variables) is the part that multiplies $\overline{\overline{N}}$.